If f(x)=((x3-3 x+6)(x-α)/2(x-α)) has range as R( set of real numbers), then α can be

Question

If f(x)=((x3-3 x+6)(x-α)/2(x-α)) has range as R( set of real numbers), then α can be (A) -1 (B) 0 (C) -2 (D) 2

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/00aac9e137bf0b794235ad668f884476-.png" /> (x3-3 x+6/2)=4 ⇒ x=-1,2 and (x3-3 x+6/2)=2 ⇒ x=1,-2 ⇒ For range to be R, α ∈[-2,2]

Q. If $f (x) = \frac{( x ^{3} - 3 x + 6 ) ( x - α )}{2 ( x - α )}$ has range as $R ($ set of real numbers), then $α$ can be

-1

0

-2

2

$\frac{x ^{3} - 3 x + 6}{2} = 4 \Rightarrow x = - 1, 2$ and $\frac{x ^{3} - 3 x + 6}{2} = 2 \Rightarrow x = 1, - 2$
$\Rightarrow$ For range to be R, $α \in [- 2, 2]$

Q. If f(x)=2(x−α)(x3−3x+6)(x−α)​ has range as R( set of real numbers), then α can be

-1

0

-2

2

2x3−3x+6​=4⇒x=−1,2 and 2x3−3x+6​=2⇒x=1,−2 ⇒ For range to be R, α∈[−2,2]

Q. If $f (x) = \frac{( x ^{3} - 3 x + 6 ) ( x - α )}{2 ( x - α )}$ has range as $R ($ set of real numbers), then $α$ can be

$\frac{x ^{3} - 3 x + 6}{2} = 4 \Rightarrow x = - 1, 2$ and $\frac{x ^{3} - 3 x + 6}{2} = 2 \Rightarrow x = 1, - 2$
$\Rightarrow$ For range to be R, $α \in [- 2, 2]$